Ex. 2 Required the length of the hyperbolic arc PLBRG, the abscissa BH = 20, the ordinate PH = 24, and the two axes AB and CD =60 and 36. Ans. 62.652.

PROBLEM XVIII.

To find the area of an hyperbola.

When the transverse axis, conjugate axis, and the abseissa, are given.

RULE. To the product of the transverse axis and abscissa, add of the square of the abscissa, and multiply the square root of the sum by 21; to this product add 4 times the square root of the product of the transverse axis and abscissa; then multiply this sum by 4 times the product of the conjugate axis and abscissa, and divide this last product by 75 times the transverse axis, the quotient will give the area of the hyperbola, nearly. *

Ex. 1. Required the area of the hyperbola PBGP, whose abscissa BH = 10, the transverse and conjugate axis AB and 30 and 18.

CD

=

Here

{21√[ (AB×BH)+&BH3]+4√(AB×RH)}×4CD×BH 75 AB

{21√[(30×10)+(§×103)]+4√(30×10)}×4×18×10

75 X 30

(21/371+4/300)×8 (21√2000+40√3)x8

√(26×7)+40√3) = (30√182+40√3)=

7

8/21 10 -X

8

25 7

151.681328, the area of the hyperbola PBGP required.

Ex. 2. What is the area of the hyperbola MBNM, the abscissa BK 25, the transverse and conjugate axis AB and CD = 50 and 30? Ans. 805.090844.

PROBLEM XIX.

To find the area of any mixtilineal figure by means of equidistant ordinates, terminated by a curve on one side, and a right line as a base on the other.

RULE. To the sum of the first and last ordinates add 4 times the sum of all the even ordinates, and twice the sum of all the

* Hutton's Mensuration.

odd ordinates, rejecting the first and last; and of this result, multiplied by the common distance of the ordinates, will give the area, very nearly. Prop. XV, B. IV.

Scholium. This rule is absolutely true for a parabola, if the ordinates are parallel to its axis. And if the distances between the ordinates is small, it is approximately true for any other curve.

Ex. 1. Required the area of an irregular figure, bounded on one side by a curve line at five equidistant ordinates, the breadths being AD=8.2, mp = 7.4 nq=9.2, or = 10.2, BC = 8.6; the length of the base AB = 39, and the common distance of the ordinates Am, mo, no, oB, each = 9.75.

T

Here [(AD+BC)+4 (mp D + or) + 2nq] × 9.75 = [8.2+8.6)+4(7.4 + 10.2) +(9.2×2)] × 9.75 = (16.8 +70.4+ 18.4) X9.75 (105.6 A ÷3)×9.75=343.2, the area of the space ADCBA required.

m

n

B

Ex. 2. Required the area of an irregular space ADqnA bounded on one side by a curve line, and divided by three equidistant ordinates perpendicular to the base An, the ordinates being AD=8, mp=6, and nq=10, the length of the base An=14, and the common distance Am, mn, each equal 7. Ans. 98.

Ex. 3. The abscissa of a parabola being 2, and the base or ordinate 12, required the area of the parabola.

Here, by taking three ordinates, of which the first and last are each nothing, the middle one being the abscissa=2, and the common distance--6; hence the area of the parabola=16 =Ans.

MENSURATION OF SOLIDS.

PROBLEM I.

To find the solidity of a sphere, spheroid, a spherical or an elliptical revoloid.

RULE.-Multiply a central conjugate section by the vertical axis, and take two-thirds of the product for the solidity. Ex. 1. What is the solidity of a sphere, whose diameter is 10 feet?

31.4159X=78.5397 to a central sec- A tion; hence, 78.5397x10x = 523.931 cubic feet the solidity.

Ex. 2. What is the solidity of a prolate spheroid, ACBD, whose vertical or fixed E axis, AB, is 10, and its revolving axis, CD, is 5?

3.14159×5×14= 19,63494 = a cen

tral conjugate section.

B

F

B:

Hence, 19.63494×10×3=130.9329, the solidity required.

Ex 3. Required the solidity

of a spherical hexagonal revoloid, BCGEDF, whose vertical axis is ten feet.

By referring to the table of Polygons. (Mensuration El. Geom.) we find the area of a hexagon, circumscribed about F a circle whose diameter is 10, is 17,320508; hence, 17,320508 X10X3=115,47005, the solidity required.

G

Ex. 4. Required the solidity of an elliptical rectangular revoloid, whose vertical axis is 48 inches, and conjugate axis is 36 inches.

36X36X48X2=41472 cubic inches. Ex. 5. What is the solidity of an elliptical rectangular revoloid, whose vertical axis is 36 inches, and whose conjugate is 48 inches? Ans. 55296 cubic inches. Ex. 6. What is the solidity of an oblate spheroid, whose revolving axis = 48, and whose conjugate or fixed axis = 36 Ans. 43429.4784 cubic inches.

inches?

Ex. 7. Required to find the solid content of the earth, supposing its circumference to be 25000 miles.

Ans. 263859375000 cubic miles.

Ex. 8. What is the solid content of a sphere, whose diameter AB 25 feet?

Ans. 8181.25 cubic feet.

Ex. 9. Required the solidity of a sphere, whose circumference is 18.6 fcet. Ans. 108.665413272 cubic feet.

PROBLEM. II.

To find the surface of a sphere or of a spherical revoloid.

RULE.-Multiply the perimeter of its central conjugate scction by the vertical axis, and the product is the whole surface. Ex. 1. What is the surface of a sphere whose diameter is 10? Ans. 31.4159x10=314.159, the surface required.

Ex. 3. Required the surface of a ball, whose diameter AB = 1 inch. Ans. 3.1416 square inches. Ex. 4. How many square inches will cover a globe of 12 inches in diameter ? Ans. 452.3904 square inches.

Ex. 5. Required the superficies of the terraqueous globe, supposing the diameter AB = 7958 miles. And if only onefourth part of its surface be dry land, and two acres sufficient to produce food for one person; how many persons can live

on the earth at one time.

Ans.

198956786.5824 sq. miles, the surface of the globe 49739196.6456 sq. miles, dry land.

15916542927 persons can live on the earth.

PROBLEM III.

To find the solidity of any segment or zone of a sphere.

RULE. To half the sum of the areas of the two bases mul tiplied by the altitude, add the solidity of a sphere whose diameter is equal to the altitude of the segment or zone. Ex. 1. What is the solidity of a spherical segment ABD, whose base is 10 and whose height oD is 2?

10×2=20.

and D'×3.14159×2×2×2=4.18876. hence 20+4.18876-24,18876 = the

solidity of the segment.

A

Ex. 2. What is the solid content of a zone EFDCE, whose height or = 30 inches, the greater diameter EF = 60 inches, and the less diameter AB =40 in- E ches? Ans. 75398.4 cubic inches.

Ex. 4. Required the solidity of the middle zone of a sphere ABDCA, the diameter of the whole sphere EF 80 inches, the height nr 64 inches.

DI

Ans. 233070.5408 cubic inches.

PROBLEM IV.

To find the convex surface of any segment or zone of a sphere, or spherical revoloid.

B

RULE.-Multiply the perimeter of a middle section of the whole sphere or revoloid, perpendicular to the vertical axis, by the height of the segment or zone.

Scholium. This is the same as the rule given in the Ele ments of Geometry for a spherical segment or zone, viz: its convex surface is there said to be equal to the height of the segment or zone, multiplied by the circumference of the sphere. The same rules as there given for segments and sectors of a sphere, will answer also for segments and sectors of right revoloids.

Ex. 1. What is the convex surface of a segment of a right revoloid, whose height is 2 feet, the perimeter of a central conjugate section of the whole revoloid being 40 feet?

40×2=80, the surface required.